So we've now talked about the energy it takes to change phase. So the energy it takes to go from a solid to a liquid and from a liquid to a gas. And we had some numbers that we used for water, 80 calories per gram going from a solid to a liquid for water, 540 calories per gram, we're going from a liquid to a gas for water. What we're going to talk about in this section is, how much energy does it take to heat up something within a phase? So if we take liquid water and we heat it from 10 degrees Celsius to 20 degrees Celsius, how much energy does that take? Or how much energy does it take to heat up the liquid water versus, say, asphalt? And as we all know, asphalt gets really, really hot on a hot day. There's two things involved in there. One is the color. Black does absorb more energy that it converts into heat, and so it will just get hotter than something that's not that color. But secondly, asphalt is really easy to heat up. It has what we call a low specific heat. And a specific heat is the amount of energy required to heat up a specific mass by a specific temperature. Now, you may recall that we talked about the definition of a calorie being the energy required to heat up one gram of water by one degree Celsius. And that is one calorie is the energy to heat one gram of water by one degree Celsius. And that's related. It's related to specific heat. So what we're going to do is we're going to specific heat. We're going to put together an equation that talks about specific heat. We're going to rearrange it to get what we call the heat equation. So specific heat is often abbreviated SH. Now, if you go on the internet, you're going to see people use all sorts of things. They're going to talk about specific heat. They're going to talk about specific heat capacity, they're going to talk about heat capacity, all sorts of words and these different things. We're just going to talk about this one term. Yes, there's lots of more complexity in here that we're ignoring, and that's OK. We just need to understand that different things require different energies to heat up by the same temperature. If you want to heat up a gram of water, it takes a different energy to heat that up than a gram of sand or a gram of asphalt. So specific heat-- we generally say it's energy. But it's the energy for every or per. And remember, per is an indicator of a division. So it's the energy for every one gram, which we're going to call a mass, and changing the temperature by one degree Celsius. We're going to call that a delta T. Now, delta in chemistry always means a change. And very specifically, it means a final minus initial. So it's whatever you end with minus whatever you started with, so a delta T is going to be a final temperature minus an initial temperature. And when I say I'm heating it up by one degree, I'm already saying I've got a change in temperature of one degree. But if I said, oh, I'm going to heat something from 10 degrees Celsius to 30 degrees Celsius, then my final T would be 30 degrees Celsius. My initial T would be 10 degrees Celsius. And my delta T would be 20 degrees Celsius. Oftentimes, we'll abbreviate this slightly and we'll say delta T is T sub f, which stands for temperature final, minus T sub i, which stands for temperature initial. That means the same thing as what we wrote above. So we're going to use this definition of specific heat as energy over mass times Delta T. And what we're going to do is we're going to change it around and solve it for mass. And I'm not going to show every step. I'm just going to show the answer here. The energy is going to be mass times specific heat times my change in temperature. And sometimes people will just say E equals mass times specific heat times Delta T. And in this class, we're going to interchange heat and energy. Because that's how we experience most energy in this class is as heat. And they're going say heat is mass times specific heat times Delta T. This is an important equation called the heat equation that we're going to use a lot when we talk about how much energy it takes to do something when we change temperature. So heat is mass times specific heat times Delta T. So for example, if I wanted to know how much energy to heat 18 grams of water from 15 degrees Celsius to 25 degrees Celsius, I could use my heat equation. And whenever there is some problem in this class-- and you know there's an equation that goes along with this, I said this for density-- you want to write down the equation first. So here, I've got a problem statement that's talking about energy. It's talking about mass. And it's talking about temperature. If I see a problem talking about energy, mass, and temperature, I know it's going to be some sort of specific heat problem. So as I write down my response, my answer, to the question, I'm going to start off by writing it down like that. Heat, well, that's what I'm trying to look for. Remember energy and heat are the same thing in this class. And so that's what I'm looking for. So I'm going to leave that here. Heat is mass. My mass is in grams. My specific heat, well, I just told you that for water the definition of a calorie was the energy it takes to heat up one gram of water by one degrees Celsius. So it turns out the specific heat of water is 1.0 calories per gram degrees Celsius. And that's going to be one of the two sets of units that we most commonly use for specific heat. The other one we're going to use is joules per gram degrees Celsius. For water, since there are 4.184 joules in a calorie, we end up with 4.184 joules per gram degrees Celsius for water. For other substances, it'll be a different number. So just because you see a specific heat somewhere doesn't mean you can put in a calorie per gram degrees Celsius or 4.184 joules per gram degrees Celsius. You've got to put in the number for the substance. But for water, the specific heat is either 4.184 joules per gram degree Celsius or 1.0 calories per gram degree Celsius, same number, different units. I need you to know those and have those available to you when you are solving problems. Because I'm not going to tell you. So if you've got a problem that's got water in it, you should know its density as 1.00 grams per milliliter. And you should know that it's specific heat is 4.184 joules per gram degree Celsius or 1.00 calories per gram degree Celsius. All right, so we've now got that. And now we're going to put in our Delta T. Remember Delta T is always final minus initial, 25 degrees Celsius minus 15 degrees Celsius. If you did that math, you get 10 degrees Celsius. And so we just have to multiply that out. And we get 18 times 1 times 10 is 180. What do we mean by units? Well, if we look back up here, grams cancels. Degrees Celsius cancels. And I'm left with calories. So I get 180 calories. For sig figs, specific heat definitely count the six figs. Mass, definitely count the sig figs. Temperature, ignore the sig figs. Remember temperature and sig figs is hard. So we're not going to worry about that too much. So we've got two sig figs in our mass. We're going to have two sig figs in our answer. So we can calculate that the heat it takes to warm up 18 grams of water by 10 degrees Celsius is 180 calories. That is how we use the heat equation. Now, just like any other algebraic equation, when we've got this equation here, we can solve for any of those variables by knowing the other ones. So if we wanted to solve for mass, if we knew heat, specific heat, and Delta T, we could solve for mass. If we wanted to solve for specific heat, if we knew heat, mass, and Delta T, we could solve for specific heat. One last thing I want to comment on is what if we cool something from let's say 50 degrees Celsius down to 20 degrees Celsius? How would we calculate Delta T? Well, Delta t is always final minus initial. And here since we're cooling down to 20 degrees Celsius, that's our final. And then we subtract 50 off of that. And we get minus 30 degrees Celsius. Now oftentimes, people have trouble with that. They're like, you can't have that minus there, that minus temperature. It's OK. It's just a temperature change. It just says minus 30 just means we lost temperature. And what that means is in order to lose temperature, you lost heat. When you plug that negative temperature into your heat equation, you're going to get a negative heat. But if you remember, the only thing negative heat means is I lost energy. And when I cool things down, I lose energy. So don't worry about having negative heat. That's OK. And negative Delta T's, that's perfectly OK as well. So here's an example we're going to try. How much energy is required to raise the temperature of 454 grams of dry bone by 10 degrees Celsius? So just like we did before, we're going to write down our heat equation and substitute in. Now, here we've already got mass, 454 grams. We've got specific heat, 0.11 calories per gram degrees Celsius. And our Delta T, we're heating it up by 10 degrees Celsius. And so we just multiply that out there. When you do that, you end up with 499 calories. Our grams cancel. Our degrees Celsius cancel. We're left with calories. But we still have to do sig figs. We've got three sig figs here. We've got two sig figs here. We're going to ignore the sig figs here. And so we only get two sig figs in our answer here. Can we write 500 calories? And the answer is no. Because that only has one sig fig. And we can't write 50 calories. Because that's a different number, even though if we wrote 50 dot, it would have the right sig figs. We can't do that. Turns out there's sometimes in life where you have to write things in scientific notation. If I write 5.0 times 10 to the 2 calories, I've got two significant figures. Because significant figures in scientific notations comes from the coefficient there. And I've got the right number. So 5.0 times 10 to the second calories, or 500 calories, to heat up 454 grams of dry bone. Here's your turn. What mass of a substance with a specific heat of 1.0 joules per gram degree Celsius can be heated by 20 degrees using 1,000 joules of energy? So again, start off by writing down that heat equation, seeing what you know and what you don't know, and see if that helps you. Now pause. All right, so let's do it. Heat equals mass times specific heat times Delta T. Do we know heat? We do, right? Heat and energy are the same. Do we know mass? No, it's what we're looking for. Do we know specific heat? We do. It's given to us, 1.0 joules per gram degrees Celsius. And do we know Delta T? It's heated by 20 degrees. That by is an indicator that we are talking about a change in temperature. And so we do know everything that we need to know in here. We need to solve this guy for mass. We've got 1,000 joules equals the mass times-- if I multiply these out, my degrees Celsius cancels. And I get 20 joules per gram. I divide both sides by 20 joules per gram, 20 joules per gram. And I get 1,000 divided by 20, which is 50 grams. It was a little hard. There's lots of units here. So when we did this part here, there's lots of units to have to write down. What I generally do instead is I solve the heat equation for my variable first. So I start with heat is mass times specific heat times Delta T. I solve this one for mass by dividing both sides by specific heat and Delta T. Cancels here. Cancels here. And I get mass is heat over specific heat times Delta T. And then I plug in my numbers. And then I don't have to do as many units running around. Because remember, we want to practice having units on all steps in our problems. And those problems that you have to write out for me, absolutely, I will require that you have units in all steps. All right, so that was our last example. Hopefully, you've got a sense of how specific heat works and how we can use it to solve different types of problems, where we're heating something up within a phase. You remember, within a phase when you heat it, temperature changes. As you change phases from our heating and cooling curves, temperature doesn't change, which is why when we changed phases, our heat equation didn't have Delta T in it. Because there's no temperature change. Thanks so much. Bye.